Question

1. Suppose we are interested in analyzing how much people spend on travel during the summer. We know that on average, people spend $1005 on travel with a population standard deviation of $1740. Suppose we take a sample of 30 people and we find that the average spent from that sample is $1850. We are interested in seeing if the amount spent on travel is increasing. a. What is the standard error? b. What is the margin of error at 90% confidence? c. Using my sample of 30, what would be the 90% confidence interval for the population mean? d. If I wanted to control my margin of error and set it to 500 at 90% confidence, what sample size would I need to take instead of the 30? e. What are the null and alternative hypotheses? f. What is the critical value at 90% confidence? g. Calculate the test statistic (using the sample of 30 and NOT the answer from part d). h. Find the p-value. i. What conclusion would be made here at the 90% confidence level?

Answer 1

e) Null Hypothesis H0:

ALTERNATIVE HYPOTHESIS Ha:

alpha=0.1

a) standard error=

b) since population standard deviation is known therefore we will use Z critical value

margin of error= Zc*S.e= 1.64*317.518= 522.54

c) M = 1850
Z = 1.64
sM = √(1740^2/30) = 317.68

μ = M ± Z(sM)
μ = 1850 ± 1.64*317.68
μ = 1850 ± 522.54
M = 1850, 90% CI [1327.46, 2372.54].

You can be 90% confident that the population mean (μ) falls between 1327.46 and 2372.54.

d) n= (1.64*1740/500)^2= 33

f) Since test is one tailed test (Right). Critical value is 1.28

g) test statistic= 1850-1005/317.518= 2.66

h) The P-Value is .003907.

The result is significant because p < .10.

Decision: REJECT H0

i) Conclusion: We have sufficient evidence to conclude that  the amount spent on travel is increasing at 90% confidence level.